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The Fundamental Theorem of Calculus

Describe the meaning of the Mean Value Theorem for Integrals.

The Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals states that a continuous function on a closed interval takes on its average value at some point in that interval. The theorem guarantees that if \(f(x)\) is continuous, a point c exists in an interval \([a,b]\) such that the value of the function at c is equal to the average value of \(f(x)\) over \([a,b].\) We state this theorem mathematically with the help of the formula for the average value of a function that we presented at the end of the preceding section.

Condensed — the full section is in OpenStax Calculus Volume 2.

Fundamental Theorem of Calculus Part 1: Integrals and Antiderivatives

As mentioned earlier, the Fundamental Theorem of Calculus is an extremely powerful theorem that establishes the relationship between differentiation and integration, and gives us a way to evaluate definite integrals without using Riemann sums or calculating areas. The theorem is comprised of two parts, the first of which, the Fundamental Theorem of Calculus, Part 1, is stated here. Part 1 establishes the relationship between differentiation and integration.

Before we delve into the proof, a couple of subtleties are worth mentioning here. First, a comment on the notation. Note that we have defined a function, \(F(x),\) as the definite integral of another function, \(f(t),\) from the point a to the point x. At first glance, this is confusing, because we have said several times that a definite integral is a number, and here it looks like it’s a function. The key here is to notice that for any particular value of x, the definite integral is a number. So the function \(F(x)\) returns a number (the value of the definite integral) for each value of x.

Second, it is worth commenting on some of the key implications of this theorem. There is a reason it is called the Fundamental Theorem of Calculus. Not only does it establish a relationship between integration and differentiation, but also it guarantees that any integrable function has an antiderivative.

Condensed — the full section is in OpenStax Calculus Volume 2.

Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem

The Fundamental Theorem of Calculus, Part 2, is perhaps the most important theorem in calculus. After tireless efforts by mathematicians for approximately 500 years, new techniques emerged that provided scientists with the necessary tools to explain many phenomena. Using calculus, astronomers could finally determine distances in space and map planetary orbits. Everyday financial problems such as calculating marginal costs or predicting total profit could now be handled with simplicity and accuracy. Engineers could calculate the bending strength of materials or the three-dimensional motion of objects. Our view of the world was forever changed with calculus.

After finding approximate areas by adding the areas of n rectangles, the application of this theorem is straightforward by comparison. It almost seems too simple that the area of an entire curved region can be calculated by just evaluating an antiderivative at the first and last endpoints of an interval.

We often see the notation \({F(x)|}_{a}^{b}\) to denote the expression \(F(b)-F(a).\) We use this vertical bar and associated limits a and b to indicate that we should evaluate the function \(F(x)\) at the upper limit (in this case, b), and subtract the value of the function \(F(x)\) evaluated at the lower limit (in this case, a).

The Fundamental Theorem of Calculus, Part 2 (also known as the evaluation theorem) states that if we can find an antiderivative for the integrand, then we can evaluate the definite integral by evaluating the antiderivative at the endpoints of the interval and subtracting.

Condensed — the full section is in OpenStax Calculus Volume 2.

Key Concepts

The Mean Value Theorem for Integrals states that for a continuous function over a closed interval, there is a value c such that \(f(c)\) equals the average value of the function. See .
The Fundamental Theorem of Calculus, Part 1 shows the relationship between the derivative and the integral. See .
The Fundamental Theorem of Calculus, Part 2 is a formula for evaluating a definite integral in terms of an antiderivative of its integrand. The total area under a curve can be found using this formula. See .

Key Equations

Mean Value Theorem for Integrals	If \(f(x)\) is continuous over an interval \([a,b],\) then there is at least one point \(c\in [a,b]\) such that \(f(c)=\frac{1}{b-a}{\int }_{a}^{b}f(x)dx.\)
Fundamental Theorem of Calculus Part 1	If \(f(x)\) is continuous over an interval \([a,b],\) and the function \(F(x)\) is defined by \(F(x)={\int }_{a}^{x}f(t)dt,\) then \({F}^{'}\text{(}x)=f(x).\)
Fundamental Theorem of Calculus Part 2	If f is continuous over the interval \([a,b]\) and \(F(x)\) is any antiderivative of \(f(x),\) then \({\int }_{a}^{b}f(x)dx=F(b)-F(a).\)

The Fundamental Theorem of Calculus

In the following exercises, use the Fundamental Theorem of Calculus, Part 1, to find each derivative.

In the following exercises, use a calculator to estimate the area under the curve by computing T₁₀, the average of the left- and right-endpoint Riemann sums using \(N=10\) rectangles. Then, using the Fundamental Theorem of Calculus, Part 2, determine the exact area.

In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2.

In the following exercises, use the evaluation theorem to express the integral as a function \(F(x).\)

In the following exercises, identify the roots of the integrand to remove absolute values, then evaluate using the Fundamental Theorem of Calculus, Part 2.

Introduction

Much of our work in Chapter has been motivated by the velocity-distance problem: if we know the instantaneous velocity function, \(v(t)\), for a moving object on a given time interval \([a,b]\), can we determine the distance it traveled on \([a,b]\)? If the velocity function is nonnegative on \([a,b]\), the area bounded by \(y = v(t)\) and the \(t\)-axis on \([a,b]\) is equal to the distance traveled. This area is also the value of the definite integral \(\int_a^b v(t) \, dt\). If the velocity is sometimes negative, the total area bounded by the velocity function still tells us distance traveled, while the net signed area tells us the object's change in position.

For instance, for the velocity function in Figure, the total distance \(D\) traveled by the moving object on \([a,b]\) is \[\begin{aligned}\end{aligned}\], and the total change in the object's position is \[\begin{aligned}\end{aligned}\]. The areas \(A_1\), \(A_2\), and \(A_3\) are each given by definite integrals, which may be computed by limits of Riemann sums (and in special circumstances by geometric formulas).

We now turn our attention to an alternate approach.

Exploration

The Fundamental Theorem of Calculus

Suppose we know the position function \(s(t)\) and the velocity function \(v(t)\) of an object moving in a straight line, and for the moment let us assume that \(v(t)\) is positive on \([a,b]\).

Then, as shown in Figure, we know two different ways to compute the distance, \(D\), the object travels: one is that \(D = s(b) - s(a)\), the object's change in position. The other is the area under the velocity curve, which is given by the definite integral, so \(D = \int_a^b v(t) \, dt\). Since both of these expressions tell us the distance traveled, it follows that they are equal, so \[\begin{aligned}\end{aligned}\].

Equation holds even when velocity is sometimes negative, because \(s(b) - s(a)\), the object's change in position, is also measured by the net signed area on \([a,b]\), which is given by \(\int_a^b v(t) \, dt\).

Perhaps the most powerful fact Equation reveals is that we can compute the integral's value if we can find a formula for \(s\). Remember, \(s\) and \(v\) are related by the fact that \(v\) is the derivative of \(s\), or equivalently that \(s\) is an antiderivative of \(v\).

Example

Determine the exact distance traveled on \([1,5]\) by an object with velocity function \(v(t) = 3t^2 + 40\) feet per second.

Solution

Since the velocity function is positive on the chosen interval, the distance traveled on \([1,5]\) is given by \[\begin{aligned}\end{aligned}\], where \(s\) is an antiderivative of \(v\). Now, the derivative of \(t^3\) is \(3t^2\) and the derivative of \(40t\) is \(40\), so it follows that \(s(t) = t^3 + 40t\) is an antiderivative of \(v\). Therefore, \[\begin{aligned}D \amp= \int_1^5 3t^2 + 40 \, dt = s(5) - s(1) \\ \amp= (5^3 + 40 \cdot 5) - (1^3 + 40\cdot 1) = 284 \ \text{feet}\end{aligned}\].

Note the key lesson of Example: to find the distance traveled, we need to compute the area under a curve, which is given by the definite integral. But to evaluate the integral, we can find an antiderivative, \(s\), of the velocity function, and then compute the total change in \(s\) on the interval. In particular, we can evaluate the integral without computing the limit of a Riemann sum.

If \(f\) is a continuous function on \([a,b]\), and \(F\) is any antiderivative of \(f\), then \(\int_a^b f(x) \, dx = F(b) - F(a)\).

Condensed — the full section is in Boelkins, Active Calculus.

Basic antiderivatives

The general problem of finding an antiderivative is difficult. In part, this is due to the fact that we are trying to undo the process of differentiating, and the undoing is much more difficult than the doing. For example, while it is evident that an antiderivative of \(f(x) = \sin(x)\) is \(F(x) = -\cos(x)\) and that an antiderivative of \(g(x) = x^2\) is \(G(x) = \frac{1}{3} x^3\), combinations of \(f\) and \(g\) can be far more complicated. Consider the functions \[\begin{aligned}\end{aligned}\].

What is involved in trying to find an antiderivative for each? From our experience with derivative rules, we know that derivatives of sums and constant multiples of basic functions are simple to execute, but derivatives involving products, quotients, and composites of familiar functions are more complicated. Therefore, it stands to reason that antidifferentiating products, quotients, and composites of basic functions may be even more challenging. We defer our study of all but the most elementary antiderivatives to later in the text.

We do note that whenever we know the derivative of a function, we have a function-derivative pair, so we also know the antiderivative of a function. For instance, since we know that \[\begin{aligned}\end{aligned}\], we also know that \(F(x) = -\cos(x)\) is an antiderivative of \(f(x) = \sin(x)\). \(F\) and \(f\) together form a function-derivative pair. Clearly, every basic derivative rule leads us to such a pair, and thus to a known antiderivative.

In Activity, we will construct a list of the basic antiderivatives we know at this time. Those rules will help us antidifferentiate sums and constant multiples of basic functions. For example, since \(-\cos(x)\) is an antiderivative of \(\sin(x)\) and \(\frac{1}{3}x^3\) is an antiderivative of \(x^2\), it follows that \[\begin{aligned}\end{aligned}\] is an antiderivative of \(f(x) = 5\sin(x) - 4x^2\), by the sum and constant multiple rules for differentiation.

Our current interest in antiderivatives is so that we can evaluate definite integrals by the Fundamental Theorem of Calculus. For that task, the constant \(C\) is irrelevant, and we usually omit it. To see why, consider the definite integral \[\begin{aligned}\end{aligned}\].

Observe that the \(C\)-values appear as opposites in the evaluation of the integral and thus do not affect the definite integral's value.

Condensed — the full section is in Boelkins, Active Calculus.

The total change theorem

Let us review three interpretations of the definite integral.

For a moving object with instantaneous velocity \(v(t)\), the object's change in position on the time interval \([a,b]\) is given by \(\int_a^b v(t) \, dt\), and whenever \(v(t) \ge 0\) on \([a,b]\), \(\int_a^b v(t) \, dt\) tells us the total distance traveled by the object on \([a,b]\).
For any continuous function \(f\), its definite integral \(\int_a^b f(x) \, dx\) represents the net signed area bounded by \(y = f(x)\) and the \(x\)-axis on \([a,b]\), where regions that lie below the \(x\)-axis have a minus sign associated with their area.
The value of a definite integral is linked to the average value of a function: for a continuous function \(f\) on \([a,b]\), its average value \(f_{\operatorname{AVG} [a,b]}\) is given by \[\begin{aligned}\end{aligned}\].

The Fundamental Theorem of Calculus now enables us to evaluate exactly (without taking a limit of Riemann sums) any definite integral for which we are able to find an antiderivative of the integrand.

A slight change in perspective allows us to gain even more insight into the meaning of the definite integral. Recall Equation, where we wrote the Fundamental Theorem of Calculus for a velocity function \(v\) with antiderivative \(V\) as \[\begin{aligned}\end{aligned}\].

If we instead replace \(V\) with \(s\) (which represents position) and replace \(v\) with \(s'\) (since velocity is the derivative of position), Equation then reads as \[\begin{aligned}\end{aligned}\].

In words, this version of the FTC tells us that the total change in an object's position function on a particular interval is given by the definite integral of the position function's derivative over that interval.

Of course, this result is not limited to only the setting of position and velocity. Writing the result in terms of a more general function \(f\), we have the Total Change Theorem.

If \(f\) is a continuously differentiable function on \([a,b]\) with derivative \(f'\), then \(f(b) - f(a) = \int_a^b f'(x) \, dx\). That is, the definite integral of the rate of change of a function on \([a,b]\) is the total change of the function itself on \([a,b]\).

differences in heights on \(f\) correspond to net signed areas bounded by \(f'\).

Condensed — the full section is in Boelkins, Active Calculus.

Summary

We can find the exact value of a definite integral without taking the limit of a Riemann sum or using a familiar area formula by finding the antiderivative of the integrand, and hence applying the Fundamental Theorem of Calculus.
The Fundamental Theorem of Calculus says that if \(f\) is a continuous function on \([a,b]\) and \(F\) is an antiderivative of \(f\), then \[\begin{aligned}\end{aligned}\]. Hence, if we can find an antiderivative for the integrand \(f\), evaluating the definite integral comes from simply computing the change in \(F\) on \([a,b]\).
A slightly different perspective on the FTC allows us to restate it as the Total Change Theorem, which says that \[\begin{aligned}\end{aligned}\], for any continuously differentiable function \(f\). This means that the definite integral of the instantaneous rate of change of a function \(f\) on an interval \([a,b]\) is equal to the total change in the function \(f\) on \([a,b]\).

The Mean Value Theorem for Integrals

Condensed — the full section is in OpenStax Calculus Volume 1.

Fundamental Theorem of Calculus Part 1: Integrals and Antiderivatives

Condensed — the full section is in OpenStax Calculus Volume 1.

Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem

Condensed — the full section is in OpenStax Calculus Volume 1.

Key Concepts

The Mean Value Theorem for Integrals states that for a continuous function over a closed interval, there is a value c such that \(f(c)\) equals the average value of the function. See .
The Fundamental Theorem of Calculus, Part 1 shows the relationship between the derivative and the integral. See .
The Fundamental Theorem of Calculus, Part 2 is a formula for evaluating a definite integral in terms of an antiderivative of its integrand. The total area under a curve can be found using this formula. See .

Key Equations

Mean Value Theorem for Integrals	If \(f(x)\) is continuous over an interval \([a,b],\) then there is at least one point \(c\in [a,b]\) such that \(f(c)=\frac{1}{b-a}{\int }_{a}^{b}f(x)dx.\)
Fundamental Theorem of Calculus Part 1	If \(f(x)\) is continuous over an interval \([a,b],\) and the function \(F(x)\) is defined by \(F(x)={\int }_{a}^{x}f(t)dt,\) then \({F}^{'}\text{(}x)=f(x).\)
Fundamental Theorem of Calculus Part 2	If f is continuous over the interval \([a,b]\) and \(F(x)\) is any antiderivative of \(f(x),\) then \({\int }_{a}^{b}f(x)dx=F(b)-F(a).\)

The Fundamental Theorem of Calculus

In the following exercises, use the Fundamental Theorem of Calculus, Part 1, to find each derivative.

In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2.

In the following exercises, use the evaluation theorem to express the integral as a function \(F(x).\)

In the following exercises, identify the roots of the integrand to remove absolute values, then evaluate using the Fundamental Theorem of Calculus, Part 2.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Find the average value of the function \(f(x)=8-2x\) over the interval \([0,4]\) and find c such that \(f(c)\) equals the average value of the function over \([0,4].\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

The formula states the mean value of \(f(x)\) is given by
\[\frac{1}{4-0}{\int }_{0}^{4}(8-2x)dx.\]
We can see in that the function represents a straight line and forms a right triangle bounded by the x- and y-axes. The area of the triangle is \(A=\frac{1}{2}(\text{base})(\text{height}).\) We have
\[A=\frac{1}{2}(4)(8)=16.\]
The average value is found by multiplying the area by \(1\text{/}(4-0).\) Thus, the average value of the function is
\[\frac{1}{4}(16)=4.\]
Set the average value equal to \(f(c)\) and solve for c.
\[\begin{array}{lll}8-2c & = & 4 \\ c & = & 2\end{array}\]
At \(c=2,f(2)=4.\)
Find the average value of the function \(f(x)=\frac{x}{2}\) over the interval \([0,6]\) and find c such that \(f(c)\) equals the average value of the function over \([0,6].\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(\text{Average value}=1.5;c=3\)
Given \({\int }_{0}^{3}{x}^{2}dx=9,\) find c such that \(f(c)\) equals the average value of \(f(x)={x}^{2}\) over \([0,3].\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

We are looking for the value of c such that
\[f(c)=\frac{1}{3-0}{\int }_{0}^{3}{x}^{2}dx=\frac{1}{3}(9)=3.\]
Replacing \(f(c)\) with c², we have
\[\begin{array}{lll}{c}^{2} & = & 3 \\ c & = & \text{\pm }\sqrt{3}.\end{array}\]
Since \(\text{-}\sqrt{3}\) is outside the interval, take only the positive value. Thus, \(c=\sqrt{3}\) ().
Given \({\int }_{0}^{3}(2{x}^{2}-1)dx=15,\) find c such that \(f(c)\) equals the average value of \(f(x)=2{x}^{2}-1\) over \([0,3].\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(c=\sqrt{3}\)
Use the to find the derivative of
\[g(x)={\int }_{1}^{x}\frac{1}{{t}^{3}+1}dt.\]

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

According to the Fundamental Theorem of Calculus, the derivative is given by
\[{g}^{'}\text{(}x)=\frac{1}{{x}^{3}+1}.\]
Use the Fundamental Theorem of Calculus, Part 1 to find the derivative of \(g(r)={\int }_{0}^{r}\sqrt{{x}^{2}+4}dx.\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\({g}^{'}\text{(}r)=\sqrt{{r}^{2}+4}\)
Let \(F(x)={\int }_{1}^{\sqrt{x}}\text{sin}\ tdt.\) Find \({F}^{'}\text{(}x).\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

Letting \(u(x)=\sqrt{x},\) we have \(F(x)={\int }_{1}^{u(x)}\text{sin}\ tdt.\) Thus, by the Fundamental Theorem of Calculus and the chain rule,
\[\begin{array}{ll} \\ {F}^{'}\text{(}x) & =\text{sin}(u(x))\frac{du}{dx} \\ & =\text{sin}(u(x))\cdot (\frac{1}{2}{x}^{-1\text{/}2}) \\ & =\frac{\text{sin}\sqrt{x}}{2\sqrt{x}}.\end{array}\]
Let \(F(x)={\int }_{1}^{{x}^{3}}\text{cos}\ tdt.\) Find \({F}^{'}\text{(}x).\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\({F}^{'}\text{(}x)=3{x}^{2}\text{cos}{x}^{3}\)
Let \(F(x)={\int }_{x}^{2x}{t}^{3}dt.\) Find \({F}^{'}\text{(}x).\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

We have \(F(x)={\int }_{x}^{2x}{t}^{3}dt.\) Both limits of integration are variable, so we need to split this into two integrals. We get
\[\begin{array}{ll} \\ F(x) & ={\int }_{x}^{2x}{t}^{3}dt \\ & ={\int }_{x}^{0}{t}^{3}dt+{\int }_{0}^{2x}{t}^{3}dt \\ & =\text{-}{\int }_{0}^{x}{t}^{3}dt+{\int }_{0}^{2x}{t}^{3}dt.\end{array}\]
Differentiating the first term, we obtain
\[\frac{d}{dx}[\text{-}{\int }_{0}^{x}{t}^{3}dt]=\text{-}{x}^{3}.\]
Differentiating the second term, we first let \(u(x)=2x.\) Then,
\[\begin{array}{ll} \\ \frac{d}{dx}[{\int }_{0}^{2x}{t}^{3}dt] & =\frac{d}{dx}[{\int }_{0}^{u(x)}{t}^{3}dt] \\ & ={(u(x))}^{3}\frac{du}{dx} \\ & ={(2x)}^{3}\cdot 2 \\ & =16{x}^{3}.\end{array}\]
Thus,
\[\begin{array}{ll} \\ \\ {F}^{'}\text{(}x) & =\frac{d}{dx}[\text{-}{\int }_{0}^{x}{t}^{3}dt]+\frac{d}{dx}[{\int }_{0}^{2x}{t}^{3}dt] \\ & =\text{-}{x}^{3}+16{x}^{3} \\ & =15{x}^{3}.\end{array}\]
Let \(F(x)={\int }_{x}^{{x}^{2}}\text{cos}\ tdt.\) Find \({F}^{'}\text{(}x).\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\({F}^{'}\text{(}x)=2x\ \text{cos}{x}^{2}-\text{cos}\ x\)
Use to evaluate
\[{\int }_{-2}^{2}({t}^{2}-4)dt.\]

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

Recall the power rule for Antiderivatives:
\[\text{If}\ y={x}^{n},\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C.\]
Use this rule to find the antiderivative of the function and then apply the theorem. We have
\[\begin{array}{ll}{\int }_{-2}^{2}({t}^{2}-4)dt & =\frac{{t}^{3}}{3}-{4t|}_{-2}^{2} \\ \\ \\ & =[\frac{{(2)}^{3}}{3}-4(2)]-[\frac{{(-2)}^{3}}{3}-4(-2)] \\ & =(\frac{8}{3}-8)-(-\frac{8}{3}+8) \\ & =\frac{8}{3}-8+\frac{8}{3}-8 \\ & =\frac{16}{3}-16 \\ & =-\frac{32}{3}.\end{array}\]
Evaluate the following integral using the Fundamental Theorem of Calculus, Part 2:
\[{\int }_{1}^{9}\frac{x-1}{\sqrt{x}}dx.\]

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

First, eliminate the radical by rewriting the integral using rational exponents. Then, separate the numerator terms by writing each one over the denominator:
\[{\int }_{1}^{9}\frac{x-1}{{x}^{1\text{/}2}}dx={\int }_{1}^{9}(\frac{x}{{x}^{1\text{/}2}}-\frac{1}{{x}^{1\text{/}2}})dx\text{.}\]
Use the properties of exponents to simplify:
\[{\int }_{1}^{9}(\frac{x}{{x}^{1\text{/}2}}-\frac{1}{{x}^{1\text{/}2}})dx={\int }_{1}^{9}({x}^{1\text{/}2}-{x}^{-1\text{/}2})dx\text{.}\]
Now, integrate using the power rule:
\[\begin{array}{ll} \\ \\ {\int }_{1}^{9}({x}^{1\text{/}2}-{x}^{-1\text{/}2})dx & ={(\frac{{x}^{3\text{/}2}}{\frac{3}{2}}-\frac{{x}^{1\text{/}2}}{\frac{1}{2}})|}_{1}^{9} \\ \\ & =[\frac{{(9)}^{3\text{/}2}}{\frac{3}{2}}-\frac{{(9)}^{1\text{/}2}}{\frac{1}{2}}]-[\frac{{(1)}^{3\text{/}2}}{\frac{3}{2}}-\frac{{(1)}^{1\text{/}2}}{\frac{1}{2}}] \\ & =[\frac{2}{3}(27)-2(3)]-[\frac{2}{3}(1)-2(1)] \\ & =18-6-\frac{2}{3}+2 \\ & =\frac{40}{3}.\end{array}\]
See .
Use to evaluate \({\int }_{1}^{2}{x}^{-4}dx.\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(\frac{7}{24}\)
James and Kathy are racing on roller skates. They race along a long, straight track, and whoever has gone the farthest after 5 sec wins a prize. If James can skate at a velocity of \(f(t)=5+2t\) ft/sec and Kathy can skate at a velocity of \(g(t)=10+\text{cos}(\frac{\pi }{2}t)\) ft/sec, who is going to win the race?

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

We need to integrate both functions over the interval \([0,5]\) and see which value is bigger. For James, we want to calculate
\[{\int }_{0}^{5}(5+2t)dt.\]
Using the power rule, we have
\[\begin{array}{ll}{\int }_{0}^{5}(5+2t)dt & ={(5t+{t}^{2})|}_{0}^{5} \\ & =(25+25)=50.\end{array}\]
Thus, James has skated 50 ft after 5 sec. Turning now to Kathy, we want to calculate
\[{\int }_{0}^{5}10+\text{cos}(\frac{\pi }{2}t)dt.\]
We know \(\text{sin}\ t\) is an antiderivative of \(\text{cos}\ t,\) so it is reasonable to expect that an antiderivative of \(\text{cos}(\frac{\pi }{2}t)\) would involve \(\text{sin}(\frac{\pi }{2}t).\) However, when we differentiate \(\text{sin}(\frac{\pi }{2}t),\) we get \(\frac{\pi }{2}\text{cos}(\frac{\pi }{2}t)\) as a result of the chain rule, so we have to account for this additional coefficient when we integrate. We obtain
\[\begin{array}{ll}{\int }_{0}^{5}10+\text{cos}(\frac{\pi }{2}t)dt & ={(10t+\frac{2}{\pi }\text{sin}(\frac{\pi }{2}t))|}_{0}^{5} \\ & =(50+\frac{2}{\pi })-(0-\frac{2}{\pi }\text{sin}\ 0) \\ & \approx 50.6.\end{array}\]
Kathy has skated approximately 50.6 ft after 5 sec. Kathy wins, but not by much!
Suppose James and Kathy have a rematch, but this time the official stops the contest after only 3 sec. Does this change the outcome?

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

Kathy still wins, but by a much larger margin: James skates 24 ft in 3 sec, but Kathy skates 29.3634 ft in 3 sec.
Consider two athletes running at variable speeds \({v}_{1}(t)\) and \({v}_{2}(t).\) The runners start and finish a race at exactly the same time. Explain why the two runners must be going the same speed at some point.
Two mountain climbers start their climb at base camp, taking two different routes, one steeper than the other, and arrive at the peak at exactly the same time. Is it necessarily true that, at some point, both climbers increased in altitude at the same rate?

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

Yes. It is implied by the Mean Value Theorem for Integrals.
To get on a certain toll road a driver has to take a card that lists the mile entrance point. The card also has a timestamp. When going to pay the toll at the exit, the driver is surprised to receive a speeding ticket along with the toll. Explain how this can happen.
Set \(F(x)={\int }_{1}^{x}(1-t)dt.\) Find \({F}^{'}\text{(}2)\) and the average value of \({F}^{\text{'}}\) over \([1,2].\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\({F}^{'}\text{(}2)=-1;\) average value of \({F}^{\text{'}}\) over \([1,2]\) is \(-1\text{/}2.\)
\(\frac{d}{dx}{\int }_{1}^{x}{e}^{\text{-}{t}^{2}}dt\)
\(\frac{d}{dx}{\int }_{1}^{x}{e}^{\text{cos}\ t}dt\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\({e}^{\text{cos}\ x}\)
\(\frac{d}{dx}{\int }_{3}^{x}\sqrt{9-{y}^{2}}dy\)
\(\frac{d}{dx}{\int }_{3}^{x}\frac{ds}{\sqrt{16-{s}^{2}}}\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(\frac{1}{\sqrt{16-{x}^{2}}}\)
\(\frac{d}{dx}{\int }_{x}^{2x}tdt\)
\(\frac{d}{dx}{\int }_{0}^{\sqrt{x}}tdt\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(\sqrt{x}\frac{d}{dx}\sqrt{x}=\frac{1}{2}\)
\(\frac{d}{dx}{\int }_{0}^{\text{sin}\ x}\sqrt{1-{t}^{2}}dt\)
\(\frac{d}{dx}{\int }_{\text{cos}\ x}^{1}\sqrt{1-{t}^{2}}dt\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(\text{-}\sqrt{1-{\text{cos}}^{2}x}\frac{d}{dx}\text{cos}\ x=|\text{sin}\ x|\text{sin}\ x\)
\(\frac{d}{dx}{\int }_{1}^{\sqrt{x}}\frac{{t}^{2}}{1+{t}^{4}}dt\)
\(\frac{d}{dx}{\int }_{1}^{{x}^{2}}\frac{\sqrt{t}}{1+t}dt\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(2x\frac{|x|}{1+{x}^{2}}\)
\(\frac{d}{dx}{\int }_{0}^{\text{ln}\ x}{e}^{t}dt\)
\(\frac{d}{dx}{\int }_{1}^{{e}^{x}}\text{ln}{u}^{2}du\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(\text{ln}({e}^{2x})\frac{d}{dx}{e}^{x}=2x{e}^{x}\)
The graph of \(y={\int }_{0}^{x}f(t)dt,\) where f is a piecewise constant function, is shown here.
1. Over which intervals is f positive? Over which intervals is it negative? Over which intervals, if any, is it equal to zero?
2. What are the maximum and minimum values of f?
3. What is the average value of f?
The graph of \(y={\int }_{0}^{x}f(t)dt,\) where f is a piecewise constant function, is shown here.
1. Over which intervals is f positive? Over which intervals is it negative? Over which intervals, if any, is it equal to zero?
2. What are the maximum and minimum values of f?
3. What is the average value of f?
ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

a. f is positive over \((1,2)\) and \((5,6),\) negative over \((0,1)\) and \((3,4),\) and zero over \((2,3)\) and \((4,5).\) b. The maximum value is 2 and the minimum is −3. c. The average value is 0.
The graph of \(y={\int }_{0}^{x}ℓ(t)dt,\) where ℓ is a piecewise linear function, is shown here.
1. Over which intervals is ℓ positive? Over which intervals is it negative? Over which, if any, is it zero?
2. Over which intervals is ℓ increasing? Over which is it decreasing? Over which, if any, is it constant?
3. What is the average value of ℓ?
The graph of \(y={\int }_{0}^{x}ℓ(t)dt,\) where ℓ is a piecewise linear function, is shown here.
1. Over which intervals is ℓ positive? Over which intervals is it negative? Over which, if any, is it zero?
2. Over which intervals is ℓ increasing? Over which is it decreasing? Over which intervals, if any, is it constant?
3. What is the average value of ℓ?
ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

a. ℓ is positive over \((0,1)\) and \((3,6),\) and negative over \((1,3).\) b. It is increasing over \((0,1)\) and \((3,5),\) and it is constant over \((1,3)\) and \((5,6).\) c. Its average value is \(\frac{1}{3}.\)
[T] \(y={x}^{2}\) over \([0,4]\)
[T] \(y={x}^{3}+6{x}^{2}+x-5\) over \([-4,2]\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\({T}_{10}=49.08,{\int }_{-4}^{3}({x}^{3}+6{x}^{2}+x-5)dx=48\)
[T] \(y=\sqrt{{x}^{3}}\) over \([0,6]\)
[T] \(y=\sqrt{x}+{x}^{2}\) over \([1,9]\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\({T}_{10}=260.836,{\int }_{1}^{9}(\sqrt{x}+{x}^{2})dx=260\)
[T] \(\int (\text{cos}\ x-\text{sin}\ x)dx\) over \([0,\pi ]\)

Symbols used here

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

x \in A,\ A \subseteq B

element of, subset
x belongs to A; every element of A is in B.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: The Fundamental Theorem of Calculus

Describe the meaning of the Mean Value Theorem for Integrals.
State the meaning of the Fundamental Theorem of Calculus, Part 1.
Use the Fundamental Theorem of Calculus, Part 1, to evaluate derivatives of integrals.
State the meaning of the Fundamental Theorem of Calculus, Part 2.
Use the Fundamental Theorem of Calculus, Part 2, to evaluate definite integrals.
Explain the relationship between differentiation and integration.
How long after she exits the aircraft does Julie reach terminal velocity?
Based on your answer to question 1, set up an expression involving one or more integrals that represents the distance Julie falls after 30 sec.

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

ନିଜେ ଚେଷ୍ଟାକରନ୍ତୁ

Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0), OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0), OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

ଅଧିକ Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests